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2 years ago

The correlation coefficient for the linear regression of the enrollment data is

Mathematics

2 answers:

Leokris [45]2 years ago

6 1

There is no enrollment data

frosja888 [35]2 years ago

3 0

The correlation coefficient for the linear regression of the enrollment data is ⇒ 0.996.

There is a VERY STRONG positive linear correlation between the x–value (years since 2000) and y–value (student enrollment).

Add another data point for 2013 — enrollment of 2250 students. What are the new statistics?

Regression equation:
y ≈ ⇒ 85.468x + ⇒ 985.029

Correlation: r ≈ ⇒ 0.988

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Trina wants to hang-glide at the same number of meters above sea level as she scuba-dived below sea level.will she fly higher th

mihalych1998 [28]

I need more information, do you know how high Jessie flew and how low under sea level she went?

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2 years ago

Liam is going on a vacation to Australia. His flight leaves at 3:15 P.M. It takes 1 hour and 35 minutes to get to the airport, a

Leni [432]

Answer:

12:40 is the anwser

Step-by-step explanation:

If his flight leaves at 3:15 and it takes and hour and 35 minutes to gett there than you would have to subtract 1 hour and 35 minutes from 3:15 so

3:15 minus 1 hour is 2:15

2:15 minus 35 minutes is 1:40

so now he wants to be and hour and 35 minutes earlier so the anwser would be

2:15 minus 1 hour is 1:15

1:15 minus 35 minutes is 12:40

8 0

1 year ago

Select all the inrational numbers from the list

snow_tiger [21]

B is one.

A < C and D are rational

Im thinking about E. It looks like a recurring decimal at first sight but I think its irrational.

8 0

3 years ago

Read 2 more answers

Rewrite the fraction in the sentence below as a percentage 21/25

Paul [167]

21/25

Change to decimal:-

Divide numerator by denominator.

21/25

21 ÷ 25 = 0.84

21/25 = 0.84

Change to a percentage:-

0.84 × 100 = 84

21/25 = 0.84 = 84%

6 0

2 years ago

Find two vectors in R2 with Euclidian Norm 1<br> whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .

Remembering,

$y_1=-3x_1\\y_2=-3x_2$

The two vectors we are looking for are:

$v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})$

5 0

2 years ago